Difference between revisions of "2012 AMC 10A Problems/Problem 2"

(Problem 2)
(Solution)
(3 intermediate revisions by 3 users not shown)
Line 1: Line 1:
== Problem 2 ==
+
== Problem ==
  
 
A square with side length 8 is cut in half, creating two congruent rectangles. What are the dimensions of one of these rectangles?
 
A square with side length 8 is cut in half, creating two congruent rectangles. What are the dimensions of one of these rectangles?
  
<math> \textbf{(A)}\ 2\ by\ 4\qquad\textbf{(B)}\ \ 2\ by\ 6\qquad\textbf{(C)}\ \ 2\ by\ 8\qquad\textbf{(D)}\ 4\ by\ 4\qquad\textbf{(E)}\ 4\ by\ 8 </math>
+
<math> \textbf{(A)}\ 2\ \text{by}\ 4\qquad\textbf{(B)}\ \ 2\ \text{by}\ 6\qquad\textbf{(C)}\ \ 2\ \text{by}\ 8\qquad\textbf{(D)}\ 4\ \text{by}\ 4\qquad\textbf{(E)}\ 4\ \text{by}\ 8 </math>
  
 
== Solution ==
 
== Solution ==
  
Cutting the square in half will bisect one pair of sides while the other side will remain unchanged. Thus, the new square is <math>\frac{8}{2}*8</math>, or  <math>\qquad\textbf{(E)}\ 4*8</math>.
+
Cutting the square in half will bisect one pair of sides while the other side will remain unchanged. Thus, the new square is <math>\frac{8}{2}</math> by <math>8</math>, or  <math>\boxed{\textbf{(E)}\ 4\ \text{by}\ 8}</math>.
 +
 
 +
== See Also ==
 +
 
 +
{{AMC10 box|year=2012|ab=A|num-b=1|num-a=3}}
 +
{{MAA Notice}}

Revision as of 14:33, 13 July 2021

Problem

A square with side length 8 is cut in half, creating two congruent rectangles. What are the dimensions of one of these rectangles?

$\textbf{(A)}\ 2\ \text{by}\ 4\qquad\textbf{(B)}\ \ 2\ \text{by}\ 6\qquad\textbf{(C)}\ \ 2\ \text{by}\ 8\qquad\textbf{(D)}\ 4\ \text{by}\ 4\qquad\textbf{(E)}\ 4\ \text{by}\ 8$

Solution

Cutting the square in half will bisect one pair of sides while the other side will remain unchanged. Thus, the new square is $\frac{8}{2}$ by $8$, or $\boxed{\textbf{(E)}\ 4\ \text{by}\ 8}$.

See Also

2012 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png